Dividing Polynomials Using Synthetic Division Synthetic Division There

Dividing Polynomials Using Synthetic Division

Synthetic Division There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x). Set divisor = 0 and 1 solve. Put answer here. x + 3 = 0 so x = - 3 -3 1 6 8 -2 up 3 these Bring number down below Addupthese line up - 3 first. Add - 9 these. Add Multiply these and these 2 +3 x - 1 This is the remainder 1 x 1 put answer above line above Put variables back in (one was of divided outthe in Sonext the Listanswer all coefficients is: (numbers in xfront x's) and in in next process so first number is one less power than constant along the top. If a term is missing, put in a 0. column original problem).

Let's try another Synthetic Division Set divisor = 0 and solve. Put answer here. 4 1 0 x 3 0 x 1 x - 4 = 0 so x = 4 0 -4 0 6 up 48 Bring number down these below Add upthese line Add up these up 4 first. Add 16 these. Add 192 Multiply these and 3 + 4 x 2 + 12 x + 48 This is these and 1 x 198 put answer remainder put answer above line Now put variables back in was above lineanswer Sonext the List all coefficients is: (numbers in (remember front of x's)one andxthe in in next divided process first number is one less in next constant alongout theintop. Don'tsoforget the 0's for missing column 3 column terms. power than original problem so x ).

Let's try a problem where we factor the polynomial completely given one of its factors. -2 4 You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. 8 -25 -50 up Bring number down below Addupthese line up - 8 first. Add 0 these. Add 50 these Multiply these and No remainder so x + 2 these 2 4 x + 0 x - 25 0 put IS a factor because it put answer above line divided in evenly above Put variables back in (one x was divided outthe in Sonext the Listanswer all coefficients is the divisor (numbers times in thefront quotient: of x's) and in in next process sothe first number is one less power You could check this by. If a term constant along top. is missing, putthan in a 0. column original problem). multiplying them out and getting original polynomial

Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this Power. Point. www. slcc. edu Shawna has kindly given permission for this resource to be downloaded from www. mathxtc. com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www. ststephens. wa. edu. au